Truss With Tension-Only Diagonals

Reaction at B
 

 

$\Sigma M_E = 0$

$7.5R_B = 3(1.5) + 4.5(3) + 2.25(3)$

$R_B = 3.3 ~ \text{kN}$
 

Section Through M-M
 

 

$\Sigma F_V = 0$

$CF_V = R_B$

$CF \sin 45^\circ = 3.3$

$CF = 3.3\sqrt{2} ~ \text{kN tension}$

$CF = 4.67 ~ \text{kN tension}$   ←   [ C ]   answer for part 1
 

At Joint B
 

 

$\Sigma F_V = 0$

$BF = R_B$

$BF = 3.3 ~ \text{kN compression}$   ←   [ B ]   answer for part 2
 

Section Through N-N
 

 

$\Sigma F_V = 0$

$DG_V + 3 = R_B$

$DG(4/5) + 3 = 3.3$

$DG = 0.375 ~ \text{kN tension}$
 

From Joint D
 

 

$\Sigma F_V = 0$

$DH + (4/5)DG = 3$

$DH + (4/5)(0.375) = 3$

$DH = 2.7 ~ \text{kN tension}$   ←   [ D ]   answer for part 3